Answer
$\ln|2x-1|-\ln|x|+C$
Work Step by Step
$I=\displaystyle \int\frac{1}{x^{2}(2-1/x)}dx$
We need a substitution such that $du$ has $x^{2}$ in the denominator
$\left[\begin{array}{llll}
u & =2-\frac{1}{x}=2-x^{-1} & & \\
& & & \\
du & =x^{-2}dx =\dfrac{dx}{x^{2}} & \Rightarrow & \dfrac{dx}{x^{2}}=du
\end{array}\right]$
$I=\displaystyle \int(\frac{1}{2-1/x})(\frac{dx}{x^{2}})=\quad $ apply the substitution
$= \displaystyle \int u^{-1}du$
special case of the power rule, $n=-1,$
$=\ln|u|+C$
bring back the variable $x$
$=\displaystyle \ln|2-\frac{1}{x}|+C$
$=\displaystyle \ln|\frac{2x-1}{x}|+C$
= $\ln|2x-1|-\ln|x|+C$
